The work-energy theorem states that the work done on a particle is equal to the change in its kinetic energy. In this case, the retardation force is doing negative work, leading to a loss of kinetic energy.
Given \( a = -2x \), where \( a \) is acceleration and \( x \) is displacement. We can write acceleration as \( a = v \frac{dv}{dx} \), so:
\( v \frac{dv}{dx} = -2x \)
\( v \ dv = -2x \ dx \)
Integrating both sides from initial to final states:
\( \int_{v_1}^{v_2} v \ dv = -2 \int_0^x x \ dx \)
\( \frac{v_2^2}{2} - \frac{v_1^2}{2} = -x^2 \)
Multiplying by mass \( m/2 \), gives
\(KE _2 - KE_1 = \frac{1}{2}m(v_2^2 - v_1^2) = -mx^2 \)
The change in kinetic energy is \( \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = -mx^2 \). Given \( m = 10 \text{ g} = 0.01 \text{ kg} = 10 \times 10^{-3} \text{ kg} = 10^{-2} \text{ kg} \).
\( \Delta KE = -mx^2 = -(10 \times 10^{-3} \text{ kg})x^2 = -10^{-2}x^2 \text{ J} \)
\( \Delta KE = -x^2 \times 10^{-2} \text{ J} \)
Comparing this with the given expression for loss of kinetic energy, \( -x^{-n} \frac{x^2 \times 10}{\frac{2}{J}} \), we find \( n = 2 \).
The value of \( n \) is \( \mathbf{2} \).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: